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Limit of a sequence

  1. Sep 15, 2005 #1
    Here is my dilemma: I am to prove that sin (n!*R*pi) has a limit, where R is a rational number. I rewrite R as a/b and I can see that whenever n=b, every subsequent term will be zero. I have tried to write this out using the definition of a limit, but I can't seem to break it down. I have been looking at these problems for a long time and I am blocked on this one.
    Thanks in advance,
  2. jcsd
  3. Sep 15, 2005 #2

    Physics Monkey

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    For a sequence [tex] a_n [/tex], we say that [tex] a_n \rightarrow L [/tex] if for every [tex] \epsilon > 0 [/tex] we can find an [tex] N(\epsilon) [/tex] such that [tex] | a_n - L | < \epsilon [/tex] for [tex] n > N(\epsilon) [/tex].

    You have guessed correctly that the limit is 0. Now I give you an [tex] \epsilon [/tex] and ask you to tell me where I should start looking so that the terms of the sequence are always closer than [tex] \epsilon [/tex] to the limit. Tell me where to look by giving me N. You've already pointed out that the terms of the sequence equal the limit beyond a certain point ...
    Last edited: Sep 15, 2005
  4. Sep 15, 2005 #3
    I think I got it. I wasn't including my b in the expansion of the n! as the spot where the sequence converges. I couln't relate the epsilon to the b or the n. It's just the algebra. That's what was giving me the headache. I had been doing the ones that are all polynomials and my brain was fried.

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